Question: $f(x) = \begin{cases} 6\sqrt{x } & \text{for} ~~~~x\gt{1} \\ 2x+4& \text{for} ~~~~ x \leq1\end{cases}$ Evaluate the definite integral. $\int^4_{-2}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $10$ (Choice B) B $16$ (Choice C) C $37$ (Choice D) D $41$
Answer: Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^4_{-2}f(x)\,dx$ $= \int^4_{1}f(x)\,dx + \int^{1}_{-2}f(x)\,dx~~~~~~$ [Why did we split at 1?] $= \int^4_{1}6\sqrt{x}\,dx + \int^1_{-2}(2x+4)\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^4_{1}6\sqrt{x}\,dx &=4x^\frac32\Bigg|^4_{{1}} \\\\ &= \left[4\cdot ( 4)^\frac32 \right] - \left[4\cdot ({1})^\frac32\right] \\\\ &= \left[32\right] -\left[4 \right] \\\\ &= {28}\end{aligned}$ The second definite integral: $\begin{aligned} \int^1_{-2}(2x+4)\,dx &=\left(x^2+4x\right)\Bigg|^1_{{-2}} \\\\ &= \left[( 1)^2 +4\cdot(1)\right] - \left[( {-2})^2 +4\cdot({-2})\right] \\\\ &= \left[5\right] -\left[-4 \right] \\\\ &= {9}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^4_{1}6\sqrt{x}\,dx + \int^1_{-2}(2x+4)\,dx$ $ = {28} + {9}$ $ = 37$ The answer $\int^4_{-2}f(x)\,dx = 37$